Dunearn Prelim 2009 Am P2 Solutions

Prelim 2009 a maths paper 2 marking scheme AMaths Paper 2 Answers 5. (a) y = 8 – x 3x2 + 4(8-x) = 36 …….(M1) 3x2 – 4x -4 = 0…….. (3x + 2)(x-2) = 0………(M1) x = - 2/3 or x = 2…….(A1) y = 26/3 or y = 6…….(A1)

Max.Mark: 100

(b) does not intersect means, the roots are not real b2 – 4ac < 0 In the equation mx – 1 = x2 – 2x + 3 x2 – (2+m)x + 4 = 0 (2 + m)2 – 4(4) < 0……………….(M2) (2 + m +4)(2 + m -4)< 0………….(B1) (m + 6) (m – 2) < 0 -6 < m < 2 ………………..(A1) Tangent means, (m+6)(m-2) = 0 …….(M1) m = -6 or m = 2 ……(A1) 6. (a) Eqn of BC: y + 3 = ½ (x+1) …….(M2) x - 2y = 5 ……………..(A1) or by y = mx + c method gr(AB ) = 2,  gr(AC) = - ½ ……..(M1) eqn of AC: y – 5 = - ½ (x-3)  x + 2y = 13 ….(A1) (b) solving the above equations: C(9,2) ……(A2) (c) gr(AD) = -2 …….(M1) Eqn of AD: 2x + y = 11 ……(B1) Solving AD and BC, D(27/5, 1/5) ……(A1) 12 5 (d) distance AD = …….(A1) 5 2. (a) radius2 = 9 + 16 = 25 …….(M1) Eqn of circle; (x-3)2 + (y+4)2 = 25 …….(M1) x2 + y2 – 6x + 8y = 0 …….(A1) (b)

x2 + y2 – 6x – 2y – 15 = 0 , center ( 3, 1) …….(M1) Gr of joining (3,1) & (2,3) = -2 ………..(M1) Eqn of chord: x – 2y = -4 ……..(A1)

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Prelim 2009 a maths paper 2 marking scheme 32  5 3 10  3 3  ……..(M1) 10  3 3 10  3 3 365  146 3 =  5  2 3 ……(A2) 100  27 1 Volume =  (32  5 3 )  (8  2 3) …….(M1) 3 226 =  8 3 …(A1) 3 4 (b) LHS = sin2x(1 + tan2x) = sin2 x(sec2 x) = tan2x = RHS ( each 1 M – total 3 M)

4. (a) width =

9. (a)  ABD =  DCB (right angles) ……(M1)  DAB =  CDB ( alternate segment theorem)…..(M1) Third angles are equal Hence, by AAA, ABD and DCB are similar …..(A1) Triangles similar to ABD are (i) DBT (ii) ADT (iii) BCT (A3- each one mark) (Order/orientation of the triangles are important here otherwise no marks) (b) (i) we have, ABD and DCB are similar AB BD AD   ……..(M1) DC CB DB Hence, DB 2  DA  CB ……(A1) (ii) From right triangle DCB, DC2 = DB2 – CB2 = DAxCB – CBxCB from (i) …..(M1) = CB x (DA – CB) …….(A1) (b) 3 cos x – 4 sin x = 1 R = 5 and  = tan-1(4/3) …..(M2) 5 cos (x + ) = 1 …..(M1) x = 25.3º, 228.4º ……(A2) dy  5  2 x  1 …..(M2) dx At x = 3, y = 6 (3,6) Eqn of tangent, x + y = 9 …….(A1)

10. (a) (i) at x = 3,

(ii) A(5,0) C(9,0) -------(B1) 5

Shaded area = ½ x 6 x 6 -  (5 x  x 2 )dx ……(M1) 3

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Prelim 2009 a maths paper 2 marking scheme 5x 2 x3  5 = 18 -    ……(M1) 3 3  2 32 = sq. units ……(A1) 3

10 (b) (i) 2[6] – 8 = 4 …..(A2) (ii) – [ 0 to 3 + 3 to 5 ] ….(M1) = - [ 6 + 4] = -10 …….(A1) (iii)

  f ( x )  me 3

2x

0

dx

5

 e 2 x  +m   from 0 to 3 = 5 …….. (M2)  2  

6

m 6 e  1  1 …….(B1) 2 2 m = 6 e 1 2 7 (a) v = 3t + 2(1 – t)2 (i) initial velocity = 2 m/s …….(A1)

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(ii) a = 6t -4(1-t) --------(M1) At t = 2, acceleration = 16 m s-2 ……..(A1) (iii) v = 0 does not have any real solution, there is no turning points Integrating velocity will give displacement …….(M1) Displacement, s = t3 –(2/3)(1 – t)3 + C …….(B1) t = 0, s = 0  c = 2/3 ……….(M1) Hence, displacement at t = 3 is 33 m ……(A1) 7 (b) f(x) = x3 – 7x2 + 4x + 12 f(-1) = 0 , hence, (x +1) is a factor …….(M1) using the division the other factors are (x -2) (x -6) …..(B2) Solution, x = -1 or x = 2 or x = 6 ……(A1) 8. (a)

x2 y = 36 …….(A1) A = 6xy + 4x2 …….(M1) Using, x2 y = 36 We get, A = 4 x 2 

216 .(detail working) ….(A1) x

dA 216   2  8 x …….(M1) dx x

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Prelim 2009 a maths paper 2 marking scheme

d 2 A 432  3  8 …….(M1) dx 2 x Solve the first diff, we get x = 3 ……(A1) Sub, in second, we get positive  A is minimum ….(M1) Min area = 108 cm2 ………(A1) 8. (b) (i) y 

ln x 3x  6

1 (3 x  6)    ln x3 dy  x  ………(M2) dx ( 3 x  6) 2 At x = 1, -1/3 …….(A1) (ii)

dy  dy  dx    …….(M1) dt  dx  x1 dt Rate of change of x at x = 1 is 0.3 units per second …..(A1)

2x  3  4 x  0 4x2 = 3 – 4x ………..(M1) (2x -1)(2x + 3) = 0 ……(B1) x = ½ or x = -3/2 ……..(A1) 1 (b) 4 3 x  log 2    5 8 3x 4 -3 = 5 ………(M1) 26x = 23 ……….(B1) x = ½ ………(A1)

3. (a)

1. y = sin x cos3x for 0 < x <  dy  sin x  3 cos 2 x sin x  cos 3 xcos x  ……..(M2) dx = cos2x [ -3sin2x +cos2x] = 0 for turning points ……(M1) x = /6, /2, 5/6 ……….(A2)

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